%!TEX program = xelatex
\documentclass[t,12pt,aspectratio=169]{beamer} % 16:9 宽屏比例，适合现代投影
%\usepackage{ctex} % 中文支持
\usepackage{amsmath, amsthm, amssymb, bm} % 数学公式与符号
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{color}

% 使用衬线字体（可选）
%\usefonttheme{serif}
%\usefonttheme{default}

% 或者更彻底地使用 Latin Modern 数学字体（需 lualatex/xelatex）
\usepackage{unicode-math}
%\setmainfont{Latin Modern Roman}
\setmathfont{Latin Modern Math}

%\usepackage{fontspec}
%\setmainfont{Arial}

%\usefonttheme{professionalfonts}      % 让我自定义字体
%\usepackage{fontspec}
%\setmainfont{Roboto}
%\usepackage{unicode-math}
%\setmathfont{Latin Modern Math}

%\setmainfont{Times New Roman}
%\setmainfont{Arial}
%\setmainfont{Fira Sans}
%\setmainfont{Linux Libertine O}
%\setmainfont{STIX Two Text}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 设置段落间距
\usepackage{setspace}
\onehalfspacing
\setlength{\parskip}{1em}  % 增加段落之间的间距为1em

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 每页增加与上面标题行的距离
\addtobeamertemplate{frametitle}{}{\vspace*{0.7em}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\setlength{\scriptspace}{2pt}  % 增加上标/下标周围的水平空白
\newcommand{\pow}[1]{^{\,#1}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usetheme{Madrid} % 主题设置（推荐简洁风格）
\usecolortheme{default} % 可选：seahorse, beaver, dolphin 等

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 信息设置
\title{Chapter 20: Automatic Proof of Identities}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 封面页
\begin{frame}
  \titlepage
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 目录页
\begin{frame}{Contents}
  \tableofcontents
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 0
%\section{INTRO.}
\begin{frame}[allowframebreaks]{intro. }

\vspace{-0.5cm}  

Special functions are back in fashion. 

Recent years have seen the development of new approaches to the theory and also of many applications of special function identities. 

Two important examples of the latter are de Branges' proof of Bieberbach's conjecture, [de Branges 85], and Apéry's proof of the irrationality of $\zeta(3)$, [van der Poorten 79]. 

{\color{red}Holonomic modules} can be used to find the differential equations satisfied by certain special functions and also to determine whether a given identity is true. 

In many cases this can be done automatically; that is, by a computer. 

In this chapter we give an introduction to the theoretical foundations of this approach, pioneered by Zeilberger and his collaborators.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 1
\section{Holonomic Functions}
\begin{frame}[allowframebreaks]{A. }

\vspace{-0.5cm}  

We will assume throughout this chapter that the coefficient field is $\mathbb{R}$. 

Thus $A_n$ will stand for $A_n(\mathbb{R})$. 

A function is {\color{red}holonomic} if it is the solution of a {\color{red}holonomic} module. 

Let $U$ be an open set of $\mathbb{R}^n$. 

If $f \in C\pow{\infty}(U)$, set
$$
\mathcal{I}_f = \{d \in A_n(\mathbb{R}) : d(f) = 0\}.
$$

Then $f$ is {\color{red}holonomic} if $A_n/\mathcal{I}_f$ is a {\color{red}holonomic} module. 

But $A_n/\mathcal{I}_f$ is isomorphic to the submodule $A_nf$ of $C\pow{\infty}(U)$ generated by $f$. 

Thus $f$ is a {\color{red}holonomic} function if and only if $A_nf$ is a {\color{red}holonomic} module.

A polynomial $f \in \mathbb{R}[x_1,\ldots,x_n]$ is a {\color{red}holonomic} function. 

Indeed, if $f$ has total degree $k$, then $\mathcal{I}_f$ contains the monomials $\delta^{\alpha}$ with $|\alpha| = k + 1$. 

Thus $A_n/\mathcal{I}_f$ is {\color{red}holonomic}, see Exercise 4.1. 

The next result is of great help in producing examples of {\color{red}holonomic} functions.

\textbf{Proposition 1.1.}
Let $f,g \in C\pow{\infty}(U)$ be {\color{red}holonomic} functions. 

Then $f+g$ and $fg$ are {\color{red}holonomic}.


\textbf{Proof:} Since $f$ and $g$ are {\color{red}holonomic} functions, the modules $A_nf$ and $A_ng$ are {\color{red}holonomic}. 

Hence $A_nf \oplus A_ng$ and its subquotient $A_n(f+g)$ are {\color{red}holonomic}, by Proposition 10.1.1. 

Thus $f+g$ is a {\color{red}holonomic} function.

Let $\mathcal{M}$ be the $A_n$-submodule of $C\pow{\infty}(U)$ generated by $\partial^{\alpha}(f)\partial^{\beta}(g)$ for all $\alpha, \beta \in \mathbb{N}^n$. 

It follows from Leibniz's rule that $A_n(fg) \subseteq \mathcal{M}$. 

Now, the multiplication map defines a homomorphism of $A_n$-modules,
$$
A_nf \otimes_{\mathbb{R}[X]} A_ng \to C\pow{\infty}(U),
$$
whose image is $\mathcal{M}$. 

Since $A_nf \otimes_{\mathbb{R}[X]} A_ng$ is {\color{red}holonomic} by Exercise 18.4.1, it follows that $A_n(fg)$ is {\color{red}holonomic} by Proposition 10.1.1.

The composition of {\color{red}holonomic} functions, however, is {\color{red}not} {\color{red}holonomic}. 

As we have seen in Ch. 5, §3, the function $\exp(\exp(x))$ does not satisfy a differential equation with polynomial coefficients. 

Hence, although the exponential function is {\color{red}holonomic}, the composite $\exp(\exp(x))$ is not {\color{red}holonomic}.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 2
\section{Hyperexponential Functions}
\begin{frame}[allowframebreaks]{B. }

\vspace{-0.5cm}  

Let $a_1, a_2, \ldots$ be a sequence of real numbers. 

It is called a {\color{red}geometric sequence}, if the sequence of ratios $a_{n+1}/a_n$ is constant. 

A natural generalization is to assume that the quotients $a_{n+1}/a_n$ are a rational function of $n$. 

These sequences are called {\color{red}hypergeometric}.

In the realm of functions, the object that corresponds to a geometric sequence is the exponential function, which satisfies: $f{\,}'/f$ is constant. 

This suggests that a function $f \in C\pow{\infty}(U)$ should be called {\color{red}hyperexponential} if $\partial_i(f)/f$ is a rational function of $x_1,\ldots,x_n$ for $i = 1,\ldots,n$. 

Note that if $q$ is a rational function in $\mathbb{R}(X)$, then $\exp(q)$ is hyperexponential. 

To produce more examples, one may use the next result.

\textbf{Proposition 2.1.}
The product of hyperexponential functions is hyperexponential.


\textbf{Proof:} Let $f,g$ be hyperexponential functions. 

By Leibniz's rule,
$$
\frac{\partial_i(fg)}{fg} = \frac{\partial_i(f)}{f} + \frac{\partial_i(g)}{g}.
$$

The result follows from the fact that the sum of rational functions is a rational function.

We will now show that hyperexponential functions are {\color{red}holonomic}. 

This is easy in one variable. 

If $f$ is hyperexponential, then $\partial(f)/f = p/q$, where $p,q$ are polynomials. 

Thus $f$ satisfies the differential equation
$$
(q\partial - p)(f) = 0.
$$

Equivalently, $f$ is a solution of the module $A_1/A_1(q\partial - p)$. 

Since $q \neq 0$ and we are in dimension 1, this module is {\color{red}holonomic}. 

Thus $f$ is {\color{red}holonomic}.

The $n$-dimensional case is very much harder. 

Suppose that $f$ is a hyperexponential function of $n$ variables. 

Then $\partial_i(f)/f = p_i/q_i$, where $p_i,q_i \in \mathbb{R}[X]$. 

Thus $f$ satisfies the system of differential equations
$$
(q_i\partial_i - p_i)(f) = 0
$$
for $1 \leq i \leq n$. 

Let $\mathcal{J} = \sum_{i=1}^{n} A_n(q_i\partial_i - p_i)$. 

Then $\mathcal{J} \subseteq \mathcal{I}_f$.

{\color{blue}Put $M = A_n/\mathcal{J}$ and let $q$ be the least common multiple of $q_1,\ldots,q_n$.} 

{\color{blue}Denote by $M[q^{-1}]$ the module $\mathbb{R}[X,q^{-1}] \otimes_{\mathbb{R}[x]} M$.} %; see Ch. 12, §5. 

Every element of $M[q^{-1}]$ can be written in the form $q^{-k} \otimes u$, where $u \in M$. 

Recall that the action of $\partial_i$ on $q^{-k} \otimes u \in M[q^{-1}]$ is defined by
$$
\partial_i(q^{-k} \otimes u) = -kq^{-k-1}\partial_i(q) \otimes u + q^{-k} \otimes \partial_iu.
$$

\textbf{Theorem 2.2.}
If $M[q^{-1}]$ is {\color{red}holonomic}, then $f$ is {\color{red}holonomic}.


\textbf{Proof:} Consider the map
$$
\phi : A_n/\mathcal{I}_f \to \mathbb{R}[X,q^{-1}] \otimes A_n/\mathcal{I}_f
$$
defined by $\phi(d+\mathcal{I}_f) = 1 \otimes (d+\mathcal{I}_f)$. 

By Proposition 12.5.1, the element $d+\mathcal{I}_f$ is in the kernel of $\phi$ if and only if $q^kd \in \mathcal{I}_f$ for some positive integer $k$. 

This happens if $q^kd(f) = 0$. 

Since $q$ is a polynomial, we must have that $d(f) = 0$. 

Hence $d \in \mathcal{I}_f$. 

Summing up: the homomorphism $\phi$ is injective.

Now, since $\mathcal{J} \subseteq \mathcal{I}_f$, it follows that $A_n/\mathcal{I}_f$ is a homomorphic image of $M$. 

Thus $\mathbb{R}[X,q^{-1}] \otimes A_n/\mathcal{I}_f$ is a homomorphic image of $M[q^{-1}]$, by Theorem 12.4.6. 

Hence $\mathbb{R}[X,q^{-1}] \otimes A_n/\mathcal{I}_f$ is {\color{red}holonomic} over $A_n$. 

But we have seen that $A_n/\mathcal{I}_f$ is a submodule of $\mathbb{R}[X,q^{-1}] \otimes A_n/\mathcal{I}_f$, thus it is itself {\color{red}holonomic}. 

Hence $f$ is a {\color{red}holonomic} function.

Let us prove that $M[q^{-1}]$ is a {\color{red}holonomic} $A_n$-module. 

We begin with a technical lemma.


\textbf{Lemma 2.3.}
Let $s = \max\{\deg(p_i) : 1 \leq i \leq n\} + \deg(q)$. 

If $d$ is an operator of $A_n$ of degree $k$, then
$$
q^kd \equiv g \pmod{\mathcal{J}}
$$
where $g$ is a polynomial of degree $\leq ks$.



\textbf{Proof:} The operator $d$ is a finite linear combination of monomials $x^{\alpha}\partial^{\beta}$ and $q$ commutes with $x^{\alpha}$. 

Thus we have only to prove the following statement: if $\beta \in \mathbb{N}^n$ and $|\beta| = k$, then $q^k\partial^{\beta} \equiv g \pmod{\mathcal{J}}$ where $g \in \mathbb{R}[X]$ has degree $\leq ks$.

We will proceed by induction on $k$. 

If $k = 1$, then $\partial^{\beta} = \partial_i$ for some $i = 1,\ldots,n$. 

Note that since $q$ is the least common multiple of $q_1,\ldots,q_n$, it follows that $q/q_i$ is a polynomial. 

Thus,
$$
q\partial_i \equiv p_i(q/q_i) \pmod{\mathcal{J}}.
$$
and $\deg(p_i(q/q_i)) \leq \deg(p_i) + \deg(q/q_i) \leq s$.

Assume that the result holds for $k-1$ and let us prove that $q^k\partial^{\beta} \in \mathbb{R}[X] + \mathcal{J}$ when $|\beta| = k \geq 2$. 

Since $|\beta| > 0$, it follows that $\beta_i \neq 0$ for some $i = 1,\ldots,n$. 

Now
\begin{equation}
q^k\partial^{\beta} = q^k\partial_i(\partial^{\beta-\epsilon_i}).
\tag{2.4}
\end{equation}

But
$$
q^k\partial_i = \partial_i \cdot q^k - kq^{k-1}\partial_i(q).
$$

By the induction hypothesis there exists $g \in \mathbb{R}[X]$ of degree $\leq (k-1)s$ such that $q^{k-1}\partial^{\beta-\epsilon_i} \equiv g \pmod{\mathcal{J}}$. 

Substituting in (2.4) we obtain
$$
q^k\partial^{\beta} \equiv \partial_i \cdot q \cdot g - kg\partial_i(q) \pmod{\mathcal{J}}.
$$

But $\partial_i \cdot qg = g(q\partial_i) + \partial_i(qg)$. 

Since $q\partial_i \equiv p_i(q/q_i) \pmod{\mathcal{J}}$, we conclude that
$$
q^k\partial^{\beta} \equiv gp_i(q/q_i) + \partial_i(gq) - kg\partial_i(q) \pmod{\mathcal{J}}.
$$

The right hand side is a polynomial of degree less than or equal to
$$
\max\{(k-1)s + s, (s(k-1) + s) - 1, s(k-1) + (s-1)\} \leq sk
$$
as required.

The proof of the theorem follows the argument already used in Theorem 10.3.2 and Ch. 12, §5.

\textbf{Theorem 2.5.}
The $A_n$-module $M[q^{-1}]$ is {\color{red}holonomic}.


\textbf{Proof:} Let $m = \deg(q)$ and $\Gamma$ be the good filtration of $M$ induced by the Bernstein filtration of $A_n$. 

Define
$$
\Omega_k = \{q^{-k} \otimes u : u \in \Gamma_{(m+1)k}\}.
$$

This is a filtration of $M[q^{-1}]$ as shown in Theorem 12.5.4. 

By Lemma 2.3,
$$
q^{-k} \otimes u = q^{-k(m+2)} \otimes q^{k(m+1)}u = q^{-k(m+2)} \otimes g,
$$

where $g$ is a polynomial of degree $\leq s(m+1)k$. 

Thus, as a real vector space,
$$
\dim \Omega_k \leq \binom{s(m+1)k + n}{n}.
$$

By Lemma 10.3.1, it follows that $M[q^{-1}]$ is {\color{red}holonomic}.

Putting together Theorems 2.2 and 2.5 we have the required result.

\textbf{Theorem 2.6.}
A hyperexponential function in $n$ variables is {\color{red}holonomic}.


A more general result can be found in [Takayama 92, Appendix]. 

In the next section we explain the theoretical foundations of an algorithm which calculates the differential equation satisfied by a given definite integral with parameters. 

As an example, we calculate the integral of a hyperexponential function.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 3
\section{The Method}
\begin{frame}[allowframebreaks]{C. }

\vspace{-0.5cm}  

Denote by $x,y$ the coordinate functions of $\mathbb{R}^2$. 

Let $U = (a,b) \times (-\infty,\infty)$ and $f$ be a function in $C\pow{\infty}(U)$ which satisfies
\begin{equation}
\lim_{y \to \pm \infty} x^{\alpha}\partial_y^{\beta}f(x,y) = 0
\label{tag-3-1}
\end{equation}

for all $\alpha,\beta \in \mathbb{N}^2$. 

This implies that if $P \in A_2$, then
$$
\int_{-\infty}^{\infty} P(f)dy < \infty.
$$
for every $x \in (a,b)$. 

In particular this is the case for $P = 1$. 

Put
$$
R(x) = \int_{-\infty}^{+\infty} f(x,y)dy.
$$

We will describe an algorithm which finds a differential equation satisfied by $R(x)$, when $f$ is a given {\color{red}holonomic} function.

Let $M = A_2/\mathcal{I}_f$ and $\pi : \mathbb{R}^2 \to \mathbb{R}$ be the projection on the first coordinate. 

The direct image of $M$ under $\pi$ is
$$
\pi_*M \cong M/\partial_yM \cong A_2/(\mathcal{I}_f + \partial_yA_2).
$$

If we now assume that $f$ is a {\color{red}holonomic} function, then $M$ is a {\color{red}holonomic} $A_2$-module. 

Thus $\pi_*M$ is {\color{red}holonomic} over $A_1$ by Theorem 18.2.2. 

Hence the kernel of the homomorphism
$$
A_1 \to A_2/(\mathcal{I}_f + \partial_yA_2)
$$
which maps $d \in A_1$ to $d + (\mathcal{I}_f + \partial_yA_2)$ must be non-zero. 

Thus there exists a non-zero operator $D \in A_1$ such that
$$
D = Q_1 + \partial_y \cdot Q_2
$$
where $Q_1 \in \mathcal{I}_f$ and $Q_2 \in A_2$. 

Since $Q_1(f) = 0$, we have that
\begin{equation}
D(f) = \partial_yQ_2(f).
\label{tag-3-2}
\end{equation}

Integrating the right hand side of (\ref{tag-3-2}) between $-\infty$ and $\infty$ and using the fundamental theorem of calculus,
$$
\int_{-\infty}^{\infty} \partial_yQ_2(f)dy = Q_2(f)|_{-\infty}^{+\infty} = 0.
$$

Thus, integrating the left hand side of (\ref{tag-3-2}) and using differentiation under the integral sign [Buck 56, Ch. 4, §4.4, Theorem 29], one gets
$$
0 = \int_{-\infty}^{\infty} D(f)dy = D(R).
$$

Summing up: $R(x)$ satisfies the differential equation $D(R) = 0$.

We will apply the method to an example, borrowed from [Almkvist and Zeilberger 90]. 

Let $f(x,y) = \exp(-(x/y)^2 - y^2)$ and $R(x)$ be its integral between $-\infty$ and $\infty$. 

The function $f$ is hyperexponential, hence {\color{red}holonomic} by Theorem 2.6. 

It is easy to check that it satisfies condition (\ref{tag-3-1}). 

A simple calculation shows that $f$ is a solution of the equations
$$
(y^2\partial_x + 2x)(f) = 0,
$$
$$
(y^3\partial_y + 2y^4 - 2x^2)(f) = 0.
$$

Let $L = \mathcal{I}_f + \partial_yA_2$. 

Since $y^3\partial_y = \partial_y \cdot y^3 - 3y^2$, we have that
$$
-3y^2 + 2y^4 \equiv 2x^2 \pmod{L}.
$$

We are allowed to multiply this identity by $\partial_x^3$ on the left, because $L$ is an $A_1$-submodule of $A_2$. 

Doing this and using the identity $y^2\partial_x \equiv -2x \pmod{\mathcal{J}}$, three times, we get that
$$
D = 6\partial_x \cdot x + 8\partial_x \cdot x^2 + 8x - 2\partial_x^3 \cdot x^2
$$
is in $L$. 

Now $D$ factors as
$$
(x\partial_x + 3)(\partial_x - 2)(\partial_x + 2).
$$

In this special case we may use the differential equation $D(R) = 0$ to calculate $R$. 

The general solution of this equation is of the form
$$
c_1\exp(2x) + c_2\exp(-2x) + c_3h,
$$
where $h$ satisfies the differential equation
$$
d^2h/dx^2 - 4h = Ax^{-3}
$$
for some constant $A \neq 0$.

We must now decide, among the possible solutions, which one coincides with $R(x)$. 

First of all, $h$ cannot be bounded at 0. 

But
$$
R(0) = \int_{-\infty}^{\infty} \exp(-y^2)dy = (\sqrt{\pi})^{-1};
$$
thus $c_3 = 0$. 

Furthermore, $\lim_{x \to +\infty} R(x) = 0$, hence $c_1 = 0$. 

We are left with $R(x) = c_2\exp(-2x)$, and so
$$
(\sqrt{\pi})^{-1} = R(0) = c_2.
$$

Therefore, $R(x) = (\sqrt{\pi})^{-1}\exp(-2x)$. 

This is to be taken as a mere illustration: no one would dream of calculating this particular integral in this way. 

For the easy solutions, see Exercise 4.5.

In the above example we found $D$ from the equations for $f$ by an ad hoc calculation. 

Takayama showed in [Takayama 92] that one may use Gröbner bases to determine $D$. 

If $f$ is hyperexponential, there is also an algorithm in [Almkvist and Zeilberger 90], based on Gosper's summation algorithm [Gosper 78]. 

Thus the calculation of $D$ can be done automatically.

If one tries to extend the algorithm to find $R(x)$ by solving $D(R) = 0$ like we did, the difficulties are of another order of magnitude. 

There are two main problems:

\begin{enumerate}
    \item The integration $D(R) = 0$.
    \item Checking initial conditions.
\end{enumerate}

One way to get around (1) is to settle for an algorithm to certify an identity. 

In other words, we have 'guessed' $R(x)$ by whatever method, and we want to show that our guess is correct. 

That still leaves (2). 

Since the equation $D(R) = 0$ is usually of order greater than 1, checking initial conditions can be very tricky.

However, hyperexponential functions are like hypergeometric sequences, and integrals like sums; and it turns out that these ideas can be used to certify identities of hypergeometric sums. 

In this case (2) does not present any problem and the algorithm is very effective. 

However, we are no longer dealing with modules over the Weyl algebra. 

The base ring in this case is a localization of the Weyl algebra. 

Many interesting applications of these algorithms are discussed in [Zeilberger 90], [Almkvist and Zeilberger 90] and [Cartier 92].


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 1
\section{Exercises}
\begin{frame}[allowframebreaks]{D. }

\vspace{-0.5cm}  

\textbf{Exercise 1.}
Let $\mathcal{J}_k$ be the left ideal of $A_n$ generated by the monomials $\partial^{\alpha}$ with $|\alpha| = k$. 

Show that $A_n/\mathcal{J}_k$ is a {\color{red}holonomic} $A_n$-module.

\textbf{Hint:} There exists an exact sequence,
$$
0 \to \mathcal{J}_k/\mathcal{J}_{k+1} \to A_n/\mathcal{J}_{k+1} \to A_n/\mathcal{J}_k \to 0
$$
and $\mathcal{J}_k/\mathcal{J}_{k+1} \cong K[X]$.


\newpage

\textbf{Exercise 2.}
Show that the derivative of a hyperexponential function is hyperexponential.


\newpage

\textbf{Exercise 3.}
Find an upper bound for the multiplicity of the module $A_n/\mathcal{I}_f$, when $f$ is a hyperexponential function.

\textbf{Hint:} What is the bound on the multiplicity of $M[q^{-1}]$ in Theorem 2.5?


\newpage

\textbf{Exercise 4.}
Show that the function $\sin(xy)$ is {\color{red}holonomic}, but not hyperexponential.


\newpage

\textbf{Exercise 5.}
Let $f(x,y) = \exp(-(x/y)^2 - y^2)$ and $R(x)$ be the integral of $f$ with respect to $y$ between $-\infty$ and $\infty$. 

Calculate $R(x)$ in two different ways, as follows:

\begin{enumerate}
    \item Using the substitution $t = y - x/y$ in $\int_{-\infty}^{\infty} \exp(-t^2)dt = (\sqrt{\pi})^{-1}$.
    \item Using differentiation under the integral sign to obtain the equation $R' = -2R$ and integrating it.
\end{enumerate}


\newpage

\textbf{Exercise 6.}
Let $f(x,y) = \exp(-xy^2)$ and $R(x) = \int_{-\infty}^{\infty} f(x,y)dy$. 

Find an operator $D \in A_1$ such that $D(R) = 0$.


\newpage

\textbf{Exercise 7.}
Let $\lambda_1,\ldots,\lambda_s \in \mathbb{R}^n$ and let $X$ be the $n$-tuple $(x_1,\ldots,x_n)$. 

Denote by $\lambda_i \cdot X$ the formal inner product of the two $n$-tuples. 

For $p_1,\ldots,p_s \in K[x_1,\ldots,x_n]$ put
$$
f(x) = \sum_{1}^{s} p_i(X)\exp(\lambda_i \cdot X).
$$

Show that $f$ is a {\color{red}holonomic} function.


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}


